question_answer

                The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:                                                                                                                                       

A)                 \[0.5\,\,\pi \]

B)                  \[\pi \]                

C)                 \[0.707\,\,\pi \]

D)                  zero

Correct Answer: A

Solution :

                The    displacement    equation    of   particle executing SHM is                 \[x=a\cos (\omega t+\phi )\]                                      ...(i)                 velocity, \[v=\frac{dx}{dt}=-a\,\omega \sin \,(\omega t+\phi )....(ii)\]                 and acceleration,                                 \[A=\frac{dv}{dt}=-a{{\omega }^{2}}\,\cos \,(\omega t+\phi )....(iii)\]                     Fig. (i) is a plot of Eq. (i) with \[\phi =0\]. Fig. (ii) shows Eq. (ii) also with \[\phi =0\]. Fig. (Hi) is a plot of Eq. (iii). It should be noted that in the figures the curve of v is shifted (to the left) from the curve of x by one-quarter period \[\left( \frac{1}{4}T \right)\]. Similarly, the acceleration curve of A is shifted (to the left) by \[\frac{1}{4}T\] relative to the velocity curve of v. This implies that velocity is \[{{90}^{\text{o}}}\] (0.5 \[\pi \])out of phase with the displacement and the acceleration is \[{{90}^{\text{o}}}\] (0.5 \[\pi \]) outo of phase with the velocity but \[{{180}^{\text{o}}}\] (\[\pi \]) out of phase with displacement.